Assume that $C$ is a negatively oriented, piecewise smooth, closed curve. Let $R$ be the region enclosed by $C$. Use the normal form of Green's theorem to rewrite $ \iint_R \cos(2x - 3y) + 3x - 2y^2 \, dA$ as a line integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \oint_C 2xy^2 - \dfrac{3}{2}x^2 \, dx - \dfrac{1}{3} \sin(2x - 3y) \, dy$ (Choice B) B $ \oint_C \dfrac{2}{3} y^3 - 3xy \, dx + \dfrac{1}{2}\sin(2x - 3y) \, dy$ (Choice C) C $ \oint_C \dfrac{3}{2}x^2 - 2xy^2 \, dx + \dfrac{1}{3} \sin(2x - 3y) \, dy$ (Choice D) D $ \oint_C 3xy - \dfrac{2}{3} y^3 \, dx - \dfrac{1}{2}\sin(2x - 3y) \, dy$ (Choice E) E Green's theorem is not necessarily applicable.
Answer: Assume we have a two-dimensional vector field $F(x, y) = P(x, y) \hat{\imath} + Q(x, y) \hat{\jmath}$ and a positively oriented, piecewise smooth, simple, closed curve $C$. If $R$ is the region enclosed by $C$, then the normal form of Green's theorem states: $ \oint_C -Q \, dx + P \, dy = \iint_R \left( \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} \right) dA$ If $C$ is negatively oriented, then the line integral is equal to the negative of the double integral. [How is this equivalent to the circulation form of Green's thoeorem?] Our first step should be to confirm that the given curve is compatible with using Green's theorem. Looking closely, the curve $C$ does not satisfy all the properties required for Green's theorem: the problem never specifies that $C$ is simple! Because we can't be sure whether or not the curve is simple, Green's theorem is not necessarily applicable.